# how factor a polynomial into 2 quadratics?

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Robert Wake on 15 Nov 2021
Commented: Walter Roberson on 15 Nov 2021
hi I have a 4th order polynomial like so
mypolynomial = x^4 +a*x^3 + b*x^2 + c*x + d == 0
I now want to factorise this into 2 quadratic equations, as either quadratic equation contains the natural frequency and damping of a certain flight mode of an aircraft. Is there a way to do this? I have tried the 'factor' function, and also tried
collect(mypolynomial, x^2)
however this just returns the original fourth order polynomial. The equation i wish to obtain looks something like this
(x^2 + A*x + B)(x^2 + C*x + D) = 0
where the upper case coefficients are not the same as the lower case coefficients in 'mypolynomial'.
Any help is appreciated, thanks!

John D'Errico on 15 Nov 2021
Edited: John D'Errico on 15 Nov 2021
You cannot uniquely factor a 4th degree polynomial into such a pair of quadratics. You may think that you can, but it is provably impossible to do so, and a simple counter-example is sufficient to show why.
syms x
quartic = expand((x-1)*(x-2)*(x-3)*(x-4))
quartic = But that polynomial can be trivially written as the product of two quadratic polynomials.
Q1 = expand((x-1)*(x-2))
Q1 = Q2 = expand((x-3)*(x-4))
Q2 = expand(Q1*Q2)
ans = As you can see, Q1*Q2 must yield the same fourth degree polynomial. But is there any reason I could not have done this?
Q1 = expand((x-1)*(x-3))
Q1 = Q2 = expand((x-2)*(x-4))
Q2 = expand(Q1*Q2)
ans = So we have completely different quadratic factors. There is indeed no unique way to write such a 4th degree polynomial. This is no different from saying that an integer like 210 = 2*3*5*7, can be written in any of the forms 6*35 = 10*21 = 15*14. There is no unique factorization possible. The same idea applies to polynomials.
All that you can do is to find all 4 roots, then you could pair them up in any order you wish, Whatever makes you happy. This would suffice:
xroots = solve(mypolynomial,'maxdegree',4);
Or, if the coefficients of your polynomial are all numerical values, then you can use vpasolve.
xroots = vpasolve(mypolynomial);
In some cases, your roots MAY pair naturally up into pairs of complex conjugate roots. But that still does not give you a unique factorization.
Walter Roberson on 15 Nov 2021
factor() with 'real' sounds good.

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